Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 11 - Infinite Series - 11.6 Power Series - Exercises - Page 577: 23


The series converges for all $x$ in the interval $[-1,1)$.

Work Step by Step

We apply the ratio test and by using L'Hopital rule, we have $$ \rho=\lim _{n \rightarrow \infty}\left|\frac{a_{n+1}}{a_{n}}\right|=\lim _{n \rightarrow \infty} |\frac{ x^{n+1}/\ln(n+1) }{ x^{n}/\ln n}|\\ =|x|\lim _{n \rightarrow \infty} \frac{\ln n}{\ln(n+1)}=|x|\lim _{n \rightarrow \infty} \frac{n+1}{n}=|x|. $$ Hence, the series $\Sigma_{n=2}^{\infty} x^{n}/\ln n$ converges if and only if $|x|\lt1$. That is, the interval of convergence is $(-1,1)$. Now, we check the end points: At $x=-1$, then $\Sigma_{n=2}^{\infty} x^{n}/\ln n=\Sigma_{n=2}^{\infty}(-1)^n/\ln n$, converges (Alternating Series Test). At $x=1$, then $\Sigma_{n=2}^{\infty} x^{n}/\ln n=\Sigma_{n=2}^{\infty}1/\ln n$, diverges (by the comparison test with $\frac{1}{\ln }\geq \frac{1}{n}$). Hence, the series converges for all $x$ in the interval $[-1,1)$.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.